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Fall 2019 - Problem 7

measure theory, Stone-Weierstrass

Let $J \subseteq \R$ be a compact interval, and let $\mu$ be a finite Borel measure whose support lies in $J$ . For $z \in \C \setminus J$ define

F_\mu\p{z} = \int_\R \frac{1}{z - t} \,\diff\mu\p{t}.

Prove that the mapping $\mu \mapsto F_\mu$ is one-to-one.

Solution.

Let $t_0 \in \R \setminus J$ and $\delta = d\p{t_0, J} > 0$ , since $J$ is compact. Thus, if $t \in J$ , then

\begin{aligned} \frac{1}{z - t} &= \frac{1}{\p{z - t_0} - \p{t - t_0}} \\ &= -\frac{1}{t - t_0} \frac{1}{1 - \frac{z - t_0}{t - t_0}} \\ &= -\sum_{n=0}^\infty \frac{1}{t - t_0} \p{\frac{z - t_0}{t - t_0}}^n \\ &= -\sum_{n=0}^\infty \frac{\p{z - z_0}^n}{\p{t - t_0}^{n+1}} \end{aligned}

converges locally uniformly on $\abs{z - t_0} < \delta$ . Hence,

\begin{aligned} F_\mu\p{z} &= -\int_\R \sum_{n=0}^\infty \frac{\p{z - z_0}^n}{\p{t - t_0}^{n+1}} \,\diff\mu\p{t} \\ &= \sum_{n=0}^\infty \p{-\int_\R \frac{1}{\p{t - t_0}^{n+1}} \,\diff\mu\p{t}} \p{z - z_0}^n. \end{aligned}

Hence, if $F_\mu = F_\nu$ , then by uniqueness of power series, we get

\int_\R \frac{1}{\p{t - t_0}^{n+1}} \,\diff\mu\p{t} = \int_\R \frac{1}{\p{t - t_0}^{n+1}} \,\diff\nu\p{t}

for all $n \geq 0$ . Let

A = \set{\sum_{k=1}^N \frac{c_k}{\p{t - t_0}^{n_k}} \st N \in \N,\ c_k \in \R,\ n_k \in \N}.

This is a subalgebra which separates points and vanishes nowhere on $J$ , so by Stone-Weierstrass, $A$ is dense in $C\p{J}$ . Hence, $\mu = \nu$ on a dense subset of $C\p{J}$ , so they agree on all of $C\p{J}$ . By Riesz representation, this means $\mu = \nu$ as measures on $J$ . Since $\mu$ and $\nu$ are supported on $J$ , we see that $\mu = \nu$ as measures on $\R$ , which was what we wanted to show.